特里树

实现 trie 数据结构

trie数据结构的strider讲解

class node{
    node [] node = new node[26];
    boolean flag;
    public node(){

    }
    public boolean containskey(char c){
        return node[c-'a']!=null;
    }
    public void put(char c, node n){
        node[c-'a']  = n;
    }
    public node get(char c){
        return node[c-'a'];
    }
    public void setflag() {
        this.flag = true;
    }
    public boolean getflag(){
        return this.flag;
    }
}

class trie {
    node root;
    public trie() {
        root = new node();
    }
    //will take tc : o(len) of the word
    public void insert(string word) {
        node node  = root;
        for(int i =0;i




  
  
  trie数据结构二


奋斗者的解释，以便更好理解

							

								
								

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import java.util.* ;
import java.io.*; 

class node {
    node node[] = new node[26];
    int endwith = 0;// will keep track of no. of words ending with this word
    int countprefix=0;// will keep track of no. of words starting with this word
    public node(){

    }
    public boolean containskey(char c){
        return node[c-'a']!=null;
    }
    public void put(char c, node n){
        node[c-'a'] = n;
    }
    public node get(char c){
        return node[c-'a'];
    }
    public void incrementcountprefix() {
        ++this.countprefix;
    }
    public void decrementcountprefix(){
        --this.countprefix;
    }
    public void incrementendwith(){
        ++this.endwith;
    }
    public void deleteendwith(){
        --this.endwith;
    }
    public int getcountprefix(){
        return this.countprefix;
    }
    public int getendwith(){
        return this.endwith;
    }


}
public class trie {
    node root;
    public trie() {
        // write your code here.
        root = new node();
    }

    public void insert(string word) {
        node node = root;
        for(int i =0;i




  
  
  完整字符串




// tc : o(n*l)

import java.util.* ;
import java.io.*; 

class node{
    node[] node = new node[26];
    boolean flag;
    public node(){}
    public void put(char c , node n){
        node[c-'a'] = n;
    }
    public boolean containskey(char c){
        return node[c-'a']!=null;
    }
    public node get(char c){
        return node[c-'a'];
    }
    public void setend(){
        this.flag = true;
    }
    public boolean isend(){
        return this.flag;
    }
}

class trie{
    node root;
    public trie(){
        root = new node();
    }
    public boolean checkifprefixpresent(string s){
      node node = root;
      boolean flag= true;
      for(int i = 0;i




  
  
  计算不同子串的个数


tc：在 
 中插入不同的唯一子字符串的 o(n^2)
trie数据结构


import java.util.ArrayList;

public class Solution 
{
    Node root;
    static int count;
    public Solution(){
        root = new Node();
    }

    public static int countDistinctSubstrings(String s) 
    {
        count = 0;
        //    Write your code here.
        Solution sol = new Solution();
        for(int i =0;i< s.length();i++){
            String str = "";
            for (int j =i;j< s.length();j++){
                str = str+s.charAt(j);
                sol.insert(str);
            }
        }
        return count+1;
    }
    public void insert(String s){
        Node node = root;
        for(int i =0;i< s.length();i++){
            if(!node.containsKey(s.charAt(i))){
                node.put(s.charAt(i),new Node());
                count++;
            }
            node = node.get(s.charAt(i));
        }
        node.setFlag();
    }
}
class Node {
    Node node[] = new Node[26];
    boolean flag;
    public Node(){

    }
    public boolean containsKey(char c){
        return node[c-'a']!=null;
    }
    public Node get(char c){
        return node[c-'a'];
    }
    public void put(char c, Node n){
        node[c-'a'] = n;
    }
    public void setFlag(){
        this.flag = true;
    }
    public boolean getFlag(){
        return this.flag;
    }
}